For each <em>x</em> in the interval 0 ≤ <em>x</em> ≤ 5, the shell at that point has
• radius = 5 - <em>x</em>, which is the distance from <em>x</em> to <em>x</em> = 5
• height = <em>x</em> ² + 2
• thickness = d<em>x</em>
and hence contributes a volume of 2<em>π</em> (5 - <em>x</em>) (<em>x</em> ² + 2) d<em>x</em>.
Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:
Answer:
It is not normally distributed as it has it main concentration in only one side.
Step-by-step explanation:
So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).
So, let us begin the groupings into their different classes, shall we?
Data given:
0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.
(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.
(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31
(3). 0.4 - 0.6 : there are 2 values that falls into this category.
(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.
Class interval frequency.
0.00 - 0.20. 15.
0.20 - 0.40. 2.
0.4 - 0.6. 2.
We have been given an image of a circle. We are asked to find the value of each variable.
We can see that angle b corresponds to diameter of circle. We know that the measure of an angle that is inscribed to the diameter of a circle is 90 degrees . Therefore, the value of b will be 90.
we can see that angle is an inscribed angle of arc 99 degrees. We know that measure of an inscribed angle is half the measure of inscribed arc.
Therefore, the value of a is 49.5 units.
We know that measure of all angles of a triangle is 180 degrees. The measure of 3rd angle will be half the measure of c, so we can set an equation as:
Therefore, the value of c is 81.